Re: more elegant way of doing this? (very simple)
unless you really need to always use a node test AContribution rather than name()='AContribution' It's namespace aware and likely more efficient. Are you using xsl 1 or 2? xpath 1: sum(Records/Record/*[self::AContribution|self::BContribution|self::CContribution][number()=number()]) xpath2 you can do the same, or a bit more simply: sum(Records/Record/(AContribution|BContribution|CContribution)[number()=number()]) David
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